If the sum of first 2 terms of an infinite gp is 1 and every term is twice the sum of all sucessive terms the first term is
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Suppose the 1st term is a and the common ratio is r. The infinite sum of the series is a/(1-r).
The general term is ar^n. The (infinite) sum of all successive terms is
(ar^n)r + (ar^n)r^2 + (ar^n)r^3 + ...
= (ar^n)r/(1-r).
Therefore
ar^n = 2(ar^n)r/(1-r)
1-r = 2r
r = 1/3.
We are also given that
a+ar = a(1+r) = 1.
So
a = 1/(1+r) = 3/4.
ANSWER : a = 3/4.
The general term is ar^n. The (infinite) sum of all successive terms is
(ar^n)r + (ar^n)r^2 + (ar^n)r^3 + ...
= (ar^n)r/(1-r).
Therefore
ar^n = 2(ar^n)r/(1-r)
1-r = 2r
r = 1/3.
We are also given that
a+ar = a(1+r) = 1.
So
a = 1/(1+r) = 3/4.
ANSWER : a = 3/4.
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